Question

5-sqrt9-x^2
what is the domain, the range, is it odd even or neither?

Answer 1

Domain: All real number
Range: All real numbers greater than 2
Pretty sure it's even, but let me verify.

Do you understand why the domain and range are what they are?

Answer 2

(it's even)

Answer 3

not entirely

Answer 4

The key concept in domains is understand where (if anywhere) the function is undefined. In other words, "Is there any value of x that would make this function undefined". Here, we are taking x and "squaring it". Thus, we can plug in ANY value for x and still get a number back. This means that the domain of the function is all real numbers.

I like to think of it as the set of values of over which this function is "king" (because that would be the king's domain). if any of the values are insubordinate and try to "kill the king" (make him undefined) they are clearly his enemy and not in his "domain". (I like video games).

Range is a bit harder. The best advice I can give you is to think about the lowest point on the graph of the function. Here we have:
y = 2 - x^2 which is an upside down parabola, yes?

therefore the MOST that the "y" value could ever be (the fact that we are analyzing y and not x is very important. Remember y corresponds to range, and x to domain) is 2. (in the case that x = 0). This is because of that x^2, it is going to make any negative values positive.

CORRECTION: The range of this function is All real numbers LESS THAN 2. (Sorry realizing this just now).

And this is an even function because if we take f(-x) we get f(x) back. Which is the algebraic definition of an even function.

given f(x) = 2 - x^2
f(-x) = 2 - (-x)^2
f(-x) = 2 - x^2 = f(x) => an even function.

Sorry for the long reply, but I feel this is a very important concept in the grand scheme of mathematics.

Answer 5

Thank you!!